memory allocation using runtime type information

Question

I got to know about run-time type information in c++. This can be accomplished with typeid keyword in c++.

int main()
{
        //these two where user-defined class
        Complex x;
        Point3D y;

        const type_info& s = typeid(x);

        cout<<typeid(y).name()<<endl; //works...(1)
        cout<<s.name()<<endl;         //works...(2)

        Complex * tp = new s[10];     //does not work...(3)
}

As marked in the code, I was successful in printing out types of the data objects as in (1) and (2).

Now I wish to allocate memory by using type_info/typeid. Could anyone do this? Is it even possible. I do not have any virtual functions.

Can such a feat be accomplished by any other way. I do not want to use virtual functions as it has a negative effect on code vectorization.

Answer 1

No, this is not possible. What you are looking for is a factory.

Here is an example of such a factory. Suppose you have these types:

class point_t
{
};

class complex_t : public point_t
{
};

class point3d_t : public point_t
{
};

You can define the following factory:

typedef point_t* (*makeptr)();

class point_factory
{
   static point_t* make_complex()
   {
      return new complex_t();
   }

   static point_t* make_point3d()
   {
      return new point3d_t();
   }
   static std::map<std::string, makeptr> _map;
public:
   static void init()
   {
      _map.insert(make_pair(typeid(complex_t).name(), &point_factory::make_complex));
      _map.insert(make_pair(typeid(point3d_t).name(), &point_factory::make_point3d));
   }
   static point_t* make_point(const type_info& type)
   {
      std::map<std::string, makeptr>::iterator pos = _map.find(type.name());
      if(pos != _map.end())
         return (*(pos->second))();
      return NULL;
   }
};

std::map<std::string, makeptr> point_factory::_map;

And use it like this:

point_factory::init();
point_t* p = point_factory::make_point(typeid(point3d_t));